No rational nonsingular quartic curve <span class="mathjax-formula">$\mathcal {C}_4 \subset \mathbb {P}^3$</span> can be set-theoretic complete intersection on a cubic surface

R ^ENDICONTI

del

S ^EMINARIO M ^ATEMATICO

della

U NIVERSITÀ DI P ^ADOVA

P. C. C ^RAIGHERO R. G ATTAZZO

No rational nonsingular quartic curve C

4

⊂ P

³

can be set-theoretic complete intersection on a cubic surface

Rendiconti del Seminario Matematico della Università di Padova, tome 81 (1989), p. 171-192

<http://www.numdam.org/item?id=RSMUP_1989__81__171_0>

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Nonsingular

can

be Set-Theoretic Complete Intersection

on a

Cubic Surface.

P. C. CRAIGHERO - R. GATTAZZO

(*)

RIASSUNTO - In

questo

lavoro si dimostra che in

P3, spazio proiettivo

^{di dimen-}

sione 3 sopra ^un

campo k algebricamente

chiuso di caratteristica zero,

ogni

^curva

~4,

^del

quarto

^{ordine, razionale e non}

singolare,

^non

pu6

^essere

sottoinsieme intersezione

completa

^{di alcuna}

coppia

^di

superficie (3f3,

con

3f3 superficie

cubica ed

superficie

di ordine _n,

qualunque

^{sia n.}

Introduction.

Let P3 be the

projective

space of dimension 3 over an

algebraically

closed field of characteristic zero.

The first authors who

analyzed

^the

problem

^{whether a nonsin-}

gular

^rational

quartic

^{curve can} be set-theoretic

complete

intersection

(s.t.c.i.)

^{of two surfaces were:}

1)

^L. Godeaux and D. Gallarati

(see

Introduzione in who

proved that,

^{under some}

hypotheses

^of

generality concerning

^{the kind}

of the

singularities

^{that the two surfaces have}

^~4

^{itself cannot}

be s.t.c.i. of a cubic surface

:F3

^{and a}

quartic

^surface

:F4;

2)

^{P. C.}

Graighero

ⁱⁿ

[C],

^who

proved

^that

W4’

⁼

(A4, ~,3,u, p4)

is not s.t.c.i. of any

pair

^{of surfaces}

^:F3

^and

~4;

(*)

^Indirizzo

degli

AA.: Istituto di Matematica

Applicata

dell’Università di Padova, via Belzoni 7, ^{35100 Padova.}

3)

^E.

Stagnaro

^{in who}

proved

that every

nonsingular

^ra-

tional

quartic

^curve

W,,

^{is not s.t.c.i. of} any

pair

^{of surfaces}

3E~

and

!F4;

4)

^{P. C.}

Craighero

^{and R. Gattazzo in}

[C-G],

^who

proved

^that

~~

^{is not s.t.c.i.} of any

pair

^of surfaces both of

degree

^4.

In this paper the authors _prove that

~4

^is not s.t.c.i. of any

pair

of surfaces

.9~

^and

3Fn ,

for every ^n. This is a

generalization

^{of the}

result n.

3)

^here

above,

^{and it is in the frame of a} program of research in which other authors are

presently engaged,

^{see for}

example

^{D. B.}

Jaffe,

^who concludes the Introduction of its _paper

[J] by saying:

« We

hope

^{that the}

techniques

^{of this} article will contribute to the determination of which smooth curves are set-theoretic

complete

^inter-

sections on other

types

^of surfaces in

P3,

^{for instance on}

arbitrary

ruled

surfaces,

^on

arbitrary

^cubic

surfaces,

^{or on surfaces}

having only

rational

singularities

^~.

We remark

that,

^{in the case when}

3;~

^{is a ruled surface}

(but

^not

a

cone),

^{a new method} is set up, which we think could be of interest in a more

general

^context

(see

^Remark

3, § 4).

Many

^authors

begin

^to

conjecture

^that

~4 ,

^{and even}

W,,

^{is not}

s.t.c.i. of any

pair

^{of surfaces} in P3. The aim of this _paper, as well

as the papers listed

above,

is that of

finding

^{new methods} and devices which can be

finally

^{used in}

proving

^this

conjecture,

^{or at least to}

state it for

particular degrees,

^{or for}

particular kinds,

^{of the two}

surfaces,

^{in the}

hope

^{that the}

conjecture

could be reduced to these

cases.

Actually

somewhat of this sort

happens

^{here in}

reaching

^our

result:

indeed,

^{the main Theorem}

in [S2J

^{is of}

help

ⁱⁿ

proving

^our

Prop. 1;

^the

key

^{Lemma 1 in is used in}

proving Prop. 5,

^case

3);

and

finally

the recent result of D. B. Jaffe about curves on cones

in

[J]

^{is used in the case when}

!Fa

^{is a cone}

(see Prop. 2).

Notations.

In what follows P3

denotes

the

projective

space of dimension 3

over an

algebraically closed field

of

characteristic

zero.

By

^{a cubic}

(cubic ruled)

^{surface we}

always

^{mean a reduced and} irreducible cubic

(cubic ruled)

surface in

P3; by

^{a curve W we mean a} reduced and

irreducible

^curve in P3. If -F is a surface in P3 and W a curve, with W c

-5F,

^we say that W is set-theoretic

complete

intersection

(s.t.c.i.)

^on

F,

^or

complete

intersection

(c.i.)

on /fi’ if there exists a

surface 9 such that F

n ~ _ ~,

^or

respectively

if ifi’ r1 ~ _ ~ and

~)

^{= 1. The}

expressions

^«non

singular

^rational

quartic

curve » and

« straight line(s) »

will be in the

sequel respectively

^short-

ened in « non s.r.q.c. » and « 8.1. ». Given a and a

point

^P E ~,

[P, ~]

^{will denote the}

plane joining

^P

and ~ ; given

^two

coplanar lines r, s [r, s]

^will denote the

plane joining t

^{and d.}

1. In this

paragraph

^{we examine the cubic surfaces}

/73’ containing

a non s.r.q.c. that are either non

singular

in codimension

1,

^or

cones,

concluding (see Prop. 1,

^and

Prop. 2)

^{that on} these surfaces

~4

^cannot be s.t.c.i.

LEMMA 1

(D. Gallarati).

Let -F be an irreducible

surfaces

ⁱⁿ

P3,

non

singular in

codimension

1,

^{and be a}

sur f ace in

P3 such that

where

rei

^denotes a curve on

!F,

and qi ⁼

I(CCi,

^{~’ n}

f9),

^{i =}

1,

... , t.

If

^{there exists a}

surface

^{~’ s.t.} ~ ~ ~‘’ = s’ s’ &#x3E;

0,

then there exists

a

surface

^{~ and a}

positive integer

^s

s’,

^s.t.

LEMMA 2

(E. Stagnaro). Let

^{viI be an} irreducible

(hence reduced)

monoid

sur f ace

ⁱⁿ

P3,

^{with the} vertex at 0 ⁼

(0, 0, 0 ),

^whose

degree

^is

n &#x3E;

2,

^{and which is non}

singular

ⁱⁿ codimension 1:

Let us suppose that ^we have either

with _a;

(i

⁼

1, 2) indipendent

^linear

polynomials.

^c

(a = ~

⁼

o}

be a s.l. on .~k

through 0,

and let r~ ⁼

(G

⁼

0}

^{be a}

surface of degree

^m

8UOh that

Under these

assumptions

^{have the}

following:

- in the

f irst

^caso

(*),

^{within a constant}

of k,

^{we have that}

- in the second case

(* * )

^zve have three

possibilities :

For the

proofs

of the above two Lemmas see

[S,],

pp. 139-143.

LEMMA 3

(L. Robbiano).

^{be a}

(reduced

^and

irreducible)

^sur-

face

ⁱⁿ

P3;

^{let ~ be a curve on}

!F,

^{and let F have no}

singular point

on rc.

Then, i f

^~ is s.t.c.i. on

IF, W

^{os even a c.i. on !F.}

PROOF. See

[R].

LEMMA 4. Let

~’~

^{be a cubic}

sur f ace non8ingular

ⁱⁿ codimension 1 and not ^a cone. is a s.l. which is s.t.c.i. on then there exists a

surface

^{3e such that}

PROOF. From Lemma 3 it follows that

~3

^has

necessarily

^{a sin-}

gular point

^{on ~ :}

being

^{not a} cone,

~3

^{must be a monoid} surface with

a double

point

^{D on ~. Then we can}

apply

^to

~3

and z Lemma 2

(of

^{course within a linear}

isomorphism

^{we can} suppose D ⁼

0).

^Then

the

following

cases can

happen:

a)

^{a =}

ao ,

^{with oco} irreducible of

degree 2,

^{and d = 1: we have}

in this case that

b)

^{with ao of}

degree 1,

^{and d = 2: we} have in this

case that

and, assuming

^{3Q = = we have}

c)

I with _{ai, ocz}

independent

^linear

polynomials:

^we

have in this case four

possibilities

Now,

^if

°1)

^or

c2) happens,

^{we take = =}

0)

^{or (9 = =}

0~ :

in both cases we

get

If

os)

^or

c~) happens, then, applying

^{Lemma 1 to af}

(= ~), (ai

⁼

0}

(= ~)

^{and +}

(=

^{or to J(}

(=

⁼

0} (= ~),

^{and +}

(=

respectively,

^{we can find a surface ~ such that}

PROPOSITION 1. In

p3,

^let

F3

^{be a cubic}

surface containing

^{a non}

s.r. q. c.

~4 ,

^{and let}

~3

^{b e neither}

singular

ⁱⁿ codimension

1,

^{nor a cone.}

Then

W,

^cannot be a.t.c.i. on

~3.

PROOF. Let us suppose the

contrary,

^{that is that} there exists a

surface

~, necessarily

^of

degree 4m,

^s.t.

Let 2 be the

(unique) quadric containing W,.

^{It is well known that}

9 is not a cone

(see

^e.g.

^[H],

^Ch.

^VI,

^Ex.

^6.1,

^p.

355),

^and

that,

^of

the two scrolls of s.l. on

~,

^one consists of unisecants of

~4 ,

^{and the}

others of trisecants of

W4,

^so

^W4

^{cannot be s.t.c.i. on}

~, being

^of

type (1, 3),

^{whereas a curve which is} s.t.c.i. on 9 must be of

type (d, d)

(see

^e.g.

^[H],

^Ch.

^{II, §} 6,

pp.

135-136~ .

^Let

^P,

^{be the}

algebraic cycle

of order

2,

^{which is} the residual

intersection,

^with

respect

^to

of 9 and

~3:

r2

^{can be neither an}

irreducible conic,

^{nor a}

pair

^{of distinct and}

coplanar s.l.,

^because

r2

^{would be c.i.} on 9 with the

plane

^con-

taining 1’~,

^so

that, by

^Lemma

1, applied

^{to 9}

(= eF3 (= ~)

and

F, (= lC4

^{would even be c.i.} on 9 : absurd.

r2

cannot consist of two skew

ig.l., because, by

^Lemma

1, applied

to

~3 (= eF), !2 (= ~)

^and

‘~4 (= (‘~4

^is

by (#)

^{s.t.c.i. on}

~3)

even r2

^{would be s.t.c.i. on}

F3,

^{which is}

notoriously impossible, being

a

disconnected variety (see

^e.g.

^[K],

^Ch.

VI, ~ 4,

^Ex.

4.4,

p.

199).

So it remains for

r2 only

^the

possibility

^of

being

^a

pair

^{of coin-}

cident

s.l.,

^{that is}

Now, always by

^Lemma

1, applied

^to

~3 (= ~), !2 (= ~),

^and

‘~4 (_ C1), ~

^{is s.t.c.i. on}

!F3. By

^Lemma

3, t

^must pass

through

^a

singular point

^{D of}

^~3:

^since

^!F3

^{is not a} cone, ^{D must be a} double

point

^of

~3’

^{which can be} considered ^a monoid

surface,

^non

singular

in codimension 1

by assumption,

^{of vertex D :} from Lemma

1, applied

to

~3 (= !F), !2 (_ ~), CC4 C =

^we

get

^{that t is s.t.c.i. on}

~3;

then,

from Lemma 4 we can say that there exists a surface ~ s.t.

Now,

^from

(##)

it follows that one can find a surface 9

(actually

^one

can take for T the non

reduced 3~)

^s.t.

Finally

^from

(###), (####),

^{and Lemma}

1, applied

^to

(= ~), ~,

and

W, (= 4),

^it follows that there exists a surface ;e’

(necessarily

of

degree 4)

^s.t.

~3-~= 3~"

^{which is}

absurd, according

^to

[S2] .

PROPOSITION 2.

Let:F3

^{ac cubic}

sur f ace

ⁱⁿ

P3,

^{which is a cow con-}

taining

^{a non} s.r.q.c. Then

~4

^cannot be s.t.c.i. on

~’3.

PROOF. Let us suppose the

contrary,

that is that there exists a

surface 0? s.t.

3~~

^{cannot be}

singular

in codimension one

by

^Th.

4, 3,

first case, in

[J]. ~4

must pass

through

^{the vertex V of} otherwise it would

be, by

^Lemma

3,

^{c.i. on which is absurd.} Moreover the

generic

s.l. of

!F3

^meets

~4

^{in V and in}

just

^{one further}

point: indeed,

^{if it}

met C4

^{in at least two} further distinct

points,

^{it would be a trisecant}

of

~4’

^{and it would}

belong

^to

!2,

^{so 9 c}

3;~:

^absurd

again.

^{This means}

that the

projection

^of

^~4

^{from V to a}

generic plane

^section

^~3

^of

!F3

with a

plane

^not

passing through V,

^which

presently

^{is an}

elliptic

cubic curve, is ^a birational

correspondence,

^and this would

imply

that

absurd.

Q.E.D.

2. In this

paragraph

^we examine the cubic ruled surfaces _{of gen-} eral

type,

^those

having

^{two base skew}

s.l.,

^one

double,

^{and one}

simple.

Proposition

^{3 states}

that,

^{even in this} case,

C4

cannot be s.t.c.i. on

such a surface.

LEMMA 5. Let

C4

^{be a non} s.r.q.c. in

1P3,

^let

F3

^{be a} cubic ruled

surface of general type containing ~4,

and let z£ and a be

respectively

its double and

simple

base s.l.. Then Q is a trisecant

of C4 (may

^be

tangent

at a

point

^{A and secant at a}

point B, ^{B =1=} A,

^{or even}

tangent

^to

C4

^at

a

possible f lex),

^{is a chord}

of ~4 (may

^be

tangent

^{at a}

point

which is not a

flex).

PROOF. Since d is a double s.l. on

F3,

every s.l.

meeting

o,

C4,

and

d, necessarily belongs

^to

~3’

^{so that}

~3

^{is the ruled surface of} the s.l.

meeting

^{the three curves} o,

.~, ~4. Setting ²¹

= ~,

22 = ~4’

23 = d,

^{we have that the}

degrees

^{of these curves are}

respectively

whereas the

multiplicities

^of

~1

^{= ~ and}

~3

^{= d for}

!Fa

^are respec-

tively

^{1 and 2.}

Applying

^to

^~’3

^G. Salmon’s Theorem

(see ^[G],

^p.

377)

we

get

where ri; is the number of

(distinct

^or

coincident) points

^common

to

Li

^{and 1}

i, j::;: 3,

^{i #}

j.

^From

( * )

^we

get

r12 ⁼ 4 - 2 ⁼ 2 and _r2~ ⁼ 4 - 1 ⁼

3 ;

^{whence the Lemma.}

LEMMA 6. Let

F3, d, s

^{be as} in .Lemma 5. Let a be a chord

of ~4’ intersecting transversally ^{W4 just}

^{at two distinct}

points

^{U and V.}

Let

la

^and

Iv

^{be the two}

tangent

^{8.1. to}

W4

^{at U and V}

respectively. Let

u,s set

II u

⁼

[1, Ilv

⁼

[1,

and let ou

-and 4,,

^{the two}

^rulings of ~3 passing through

^{U and V}

respectively.

^{Then we have}

PROOF. W,

^has

surely

^a

stationary point P,

^{for which there exists}

a

plane

^II

through

^{P such that = 4P. Assume A3 to be}

P3 -

II,

^{P = and the}

tangent

^to

^CC4

^{at P as the s.l.}

^{Z~ Y~.} Then,

with a suitable choice of 0 ⁼

(0, 0, 0),

^{of the unit}

point

^{ZI =}

(1, 1, 1),

and of the

parameter t,

^{we can} suppose that ^{t = oo}

corresponds

^to

Z~,

^{and that}

~4’ _

^{A3 admits one of the}

following

^two para- metric

representations:

in the first case

Zoo being

^{not a flex of}

~4’

^{in the second}

being

^{a flex}

of

~4. Working

with these

representations

^of

W,,

^one

gets easily

^the

result: we avoid

putting

^down

explicitely

^the

quite elementary

^com-

putations involved, together

with the obvious

adaptations

^{of the}

proof

when

II,

^{or V is}

^Zoo.

REMARK 1.

Using

^the

representations (**)

^{of a} rational nonsin-

gular quartic

^{in A3 c}

p3,

^{one can}

easily recognize

^{that has}

at least one affine

stationary point,

^so in any ^case has at least two distinct

stationary points.

LEMMA 7.

~’~ , .~,

^{Q be as in Zemma 5. be ac}

tangent

to one

of

^its

points U,

^{which is not a}

flex for ~4.

^Let

IIu

^{be the}

obsculating plane

^to

C4

^at

U,

and let gU be the

ruling of

^IF

through

^U.

Then we have

PROOF. One

proceeds

^{in a} way

quite

^{similar to} that followed in

proving

^{Lemma 6.}

LEMMA 8. Let ^as in Lemma 5. Let D a

point of

d r~

6’,.

^Let us assume that on the ^two

(distinct

^or

coincident) rulings of ~’3 through

D there is no

point of ~4

other than D. Then the

plane

II =

[1, D] is tangent

^to

~~

^{at D.}

PROOF. We

distinguish

^two cases,

according

^as

~4

intersects 4 at two distinct

points

^U

and V,

^{or it is}

tangent

^{to s at a}

point

^U

(which

^{is not a flex of}

~4 ) .

In the first case, the

plane

^{1’I =}

[1, D]

cannot be

tangent

^to

~4

^{neither at}

U,

^{nor at V : indeed let it be}

tangent

to

~4

^for

example

^at

U ; then, by

^Lemma

6,

^the

ruling

flu of

~’3 through

^{U would}

lay

^on

II,

^and

consequently

^we

^would

^have flu ⁼

UD, against

^our

assumption

^{on the}

rulings

^of

^~’3 through

^{D. So}

Moreover there does not exist any

point ~4’

^with

Y, D}:

indeed,

^{if it}

existed,

^{then E would}

belong

^{to one of the}

rulings

^of

through D,

and this would lead

again

^{to a} contradiction. So neces-

sarily I (D, ~4 n II ) = 2 .

In the second case one argues in ^an

analogous

way,

taking

^into

account Lemma 7.

LEMMA 9. Let

~4’

Lemma 5. Then there exists

a

point

^{D f1}

rc4

^{such that at least one}

of

^{the two}

(distinct

^or

coincident) rulings of

^~

through

D intersects on

~4

^a

point

^D.

PROOF. We shall

distinguish

^two cases,

according

^{as d meets}

~4

at three distinct

points D1, D2 ,

^{or it is}

tangent

^to

~4

^at

D1,

^and

meets

~4

^{in a further}

point D~ (possibly

coincident with

Dl, when Di

is a flex of

~4).

Case 1.

~4

^{r1 ~ _}

(Di, ^{D2, Di =1= j. By}

^Lemma

^8,

^if

for

every i

⁼

1, 2, 3, through D~

there does not pass any

ruling

^con-

taining

^a

point

^of

~4

^{different from}

Di,

each of the three distinct

planes IIa

⁼

[1,

⁼

1, 2, 3,

^{would be}

tangent respectively

^at

J9~

i ⁼

1, 2, 3,

^to

~4’

^{which is not}

possible,

because the number of

planes through a

^{which are}

tangent

^{to at}

points

^not

belonging

^{to a}

is at most

2,

^as it is easy to

verify.

^This

proofs

the Lemma in the first case.

Case 2. Let be either .~ = with .~

tangent

^to

W,

at

D1,

^{and or}

~4 r1 ~ _ ~Dl~,

^with .~ inflexional

tangent

to

W4

at its flex

Dl.

In both these

subcases,

^{II =}

[1,

^{is not tan-}

gent

^to

W4

^at

D1,

because it would contain the

tangent

^to

~4

^at

Dl ,

which is

d,

^{and d and d would be}

coplanar. Then, by

^Lemma

8,

^there

must exist a

ruling

^of

~ 3 through Dl meeting W4

^at

Dl

^{and at a further}

point Di .

^{Whence the Lemma.}

PROPOSITION 3. In

P3, for

every ^non s.r.q.c.

~4,

^and

for

every cubic ruled

sur f acce ^{~ 3} of general type containing W4,

^{we have that}

W4

^cannot

be s.t.c.i. ^on

~’3.

PROOF. Let and a be

respectively

^{the double and}

simple

^base

s.l. of Let ~1 be ^a

ruling

^of

3;~ (see

^Lemma

9) which,

^{beside a}

point DI

^E

~4 m z£,

^{contains a further}

point P1 (~ .~)

^of

~4 . Now, by

Remark

1, CC4

^{has at least two distinct}

stationary points,

^{so that there}

exists

surely

^{one of them} P which is different from

Pi,

^{and let us}

choose as

plane

^at

infinity ^Hm

^the

osculating plane

^to

W4

^{at P.}

With this

W(4a’ = ~4

ⁿ A3 admits a

parametric biregular

^represen-

tation of the

type

with

a(t), b(t), c(t) polynomials

ⁱⁿ

k[t],

^of

degree

^{4. Let}

tl

^{be the} value of the

parameter corresponding

^to

P1

⁼

P(tl). 4F§~~

^{= A3}

admits a

parametric representation

^{of the form}

where one can suppose that

i(t), m(t), n(t)

^{are three}

coprime poly-

nomials in

k[t] (in ^{(*) ~4 ’ plays}

^the role of base

curve).

^{For every}

given t E k,

^{such that}

P(t)

⁼

(a(t), ^b(t), c(t)) 0

^.~

is a

parametric representation (in

^the

parameter ~c)

for the affine

part

~ (t ) ~"’

^{of the}

^unique

^of

^~3 through P(t). (* *),

^{for t =}

tl , represents

^{a s.l.}

passing through ^Pl

⁼

P(tl)

and contained in

~3a’,

7

which

actually

is the affine

part

^{of the}

ruling

_{o, of}

F3 through Px .

Now let us suppose that there exists ^a surface such that

~’3 ~ ~ _

-

3m~4. ~

^has

necessarily degree

^4m.

Being

^{.~ a} trisecant of

~4 (possibly

^{in three}

points variously coincident),

^the

generic ruling

^{of F}

is a s.l.

intersecting transversally ~4

ⁱⁿ

just

^one

point:

^{this means}

that,

^for

generic t, ~(t)

intersects

~4 just

ⁱⁿ

P(t),

^{which in}

(* * )

^is

obtained for the value u ⁼ 0 of the

parameter

^{u. Now we}

have,

^for

every

such t,

which means that

a(t)

intersects 9 at

P(t)

^with

I(P(t), a(t)

ⁿ

~) ==

= 4m ⁼⁼

deg (9).

This

implies that,

^if

in

k[t, u]

^{we have the}

following

with suitable

A(t)

^E

k[t]. Putting

ⁱⁿ

( * * * )

^{t =}

t1,

^we

get

Now,

^{if it were}

A(ti)

^~

0,

this would mean that the

ruling

y, inter- sects 9 at

Pl

⁼

P(t1)

^with

multiplicity 4m,

^{and this would}

prevent

~1 from

having

^{on it another}

point

^of

CC4 c ~: contradiction,

^because

we know that

D1 E /11.

^{So it is =}

^0;

^{but then we have ~l c}

^~,

hence ~1 c 9 ^r1

~’3 = ~4:

^absurd.

Q.E.D.

3. In this

paragraph

^we examine the case in which

~’3

^{is a cubic}

ruled surface of

Cayley’s type containing

^{a non} s.r.q.c. The main statements are

Prop.

^{4 and}

Prop.

^5.

Let us remember that

~’3

^{has a} base double that there exists

just

^one

plane

^II*

through

^{.~ which osculates}

~’3 along d,

^{that is s.t.}

=

3.~;

and that there exists

just

^one

point

^{D E d s.t. the}

only

s.l. of

~’3 through

^{D is d}

(D

^{is the}

uniplanar

^double

point

^of

with H* as

principal plane).

REMARK 2. The

proofs

^{of the}

statements

contained in this Remark

are not

given explicitely, being

either well

known,

^{or of not difficult}

verification. Let

~4

^{be a non} s.r.q.c. in

P3,

^{and lot 4 be a} s.l. which is an

ordinary

^{chord of}

~4’ intersecting

^it

transversally just

^{at two}

distinct

points

A. and B. Let P be a

generic point

^of

~4:

^The

plane

H =

[~, P]

intersects on

~4’

^beside

A, B and P,

^{a further}

point

^~’’.

1)

^The

correspondence

such that ⁼ P’ is birational and

biregular:

^since

~4 !:::!. P1,

^{S2 is}

a bilinear involution.

2)

^Let

Pl

^and

P2

^{be the two}

(necessarily distinct)

^fixed

points

of ,52. Moreover let ~ be the ruled surface whose

generic ruling

^is

the s.l.

[P, Q(P)] (such

surface has

obviously

^{as base curve both}

~4

and

~).

^{Then the two}

tangents

o, and ~2 ^to

respectively

^at

P,

and

P2 , belong

^{to 1’.}

3)

^Let

/~

^and

/~

^{be the}

tangents

^to

~4

^{at A and}

B, respectively.

Suppose

^that

Ih

⁼

[.t~, ~]

^is

osculating CC4

^{at A : then}

S2(A)

⁼

A,

and so,

by 2), IA belongs

^{to 3Q}

(similarly ^{S~(B) = B,}

^and

^{t’B c.4,}

^if

IIB

⁼

[.tB, ~]

^is

osculating CC4

^at

B).

4)

^If

/~

and

IB, (see 3)),

^{are not}

^coplanar,

and neither

IIA,

^y

nor

lIB (see 3))

^is

osculating CC4 respectively

^{at A or at}

B,

then every

plane

^II

through t.

intersects

on W,

^{at least one}

point ^{P =1=} A,

^B.

LEMMA 10. Let

~4

^{be a non} s.r.q.c.

CC4

^{c P3. Let}

~3

^{be a} cubic ruled

surface of Cayl,ey’s type containing ~4 .

Let d be the base s. t.

of J~7,

^and

suppose that d is ^an

ordinary

^chord

of ~4 intersecting

^it

transversally just

^{at A and}

B, ^{A 0}

^{B. Then}

necessarly

^{the two s.Z.}

IA

^{and which}

are

tangent

^to

CC 4 respectively

^{at A and}

B,

^{must be}

coplanar.

PROOF.

Suppose

^that

/~

^and

/~

^{were not}

coplanar.

^{In our situa-}

tion,

^the ruled surface ~f of Remark 2 is

just

^{with + = ~. Let}

II _{be any}

plane through d,

^{and let us} prove that II cannot be oscu-

lating ~’3 along d,

that is it cannot be ⁼ 3.~: this shall be a

contradiction,

^so

/~

^and

/~

^{must be}

coplanar.

^{Let us choose II =}

First suppose

I(A,

⁼

2 ;

^then

I (B, ~4 n II~)

^{= 1}

(it

^cannot

be

2,

^otherwise

/~

^and

IB

would be both contained in

IIA,

^{that is}

they

^{would be}

coplanar) ;

^so it exists and this

means that it cannot be ⁼ 3.~.

Now _suppose

7(~~4~77~)=3; then, by

^Remark

2, 3 ),

^{it fol-}

lows that

/ c

^and

again

^{it cannot be = 3~.} Same argu-

ment when II =

lIB. Finally

^{if then =}

=

7(-By

⁼

1,

^so it exists at least one

point

^and

d,

^{and this}

again prevents

^{II from}

osculating ~ 3 along

^.~.

LEMMA 11. Let

~4

^{be a non} s.r.q.c. in ~’3. Let A be a

point o f ~4

which is not a

flex o f ~4

^{and d the}

tangent

^{to A. Let us} sup- pose that

~4 r1.~ _ {~4.}.

..Let B be any

point o f

^.~

different f rom

^A.

T hen

through

^{B there}

passes at

^{least one}

s.l., di f f erent

^which

either two distinct

points,

^{or it is}

tangent to

^suit- able

point (different A).

PROOF.

Let y

^{be a}

plane

transversal

to d,

^{and let}

~4

- y be the

projection

^of

^~4

^{from B to} y, and

~4

^the

image

^of

~4

under ffJB;

it is easy ^{to see} that

~4

^{is a}

^quartic

^curve

of y

^and

~4 -+ ~~

^is

a birational

(and regular) correspondence,

^{so that}

~4

^{is a rational}

quartic

^curve of y. Let ^{us set} A’=

Y B(A):

^{it is}

quite simple

^{also to}

see that A’ is an

ordinary

cusp of

~~,

^which

^implies

^that

~4

^itself

has at least one further

multiple point D, ^{D -::;6}

^A’.

Now,

^if

through

^B

there did not pass any chord ^or

tangent

^to

~4

^different

from d,

every

point

^of

~4

^different from A’ would be of course a

simple point

^of

~~ :

absurd.

LEMMA 12. In ~3 there does not exist any cubic ruled

~3 o f type containing a

^{non s.r.q.c.}

~4,

^and

having acs

^{its double}

base

tangent to point A,

^{not a}

flex o f ~4 ,

^and

such that

~4~~= ~A~.

PROOF. Let us suppose that such ^a surface

~3

^{exists and deduce}

a contradiction. Let P be a

generic point

^of

~4’

^{and let us considere}

77p=[~7~].

^{We have}

lIP ~ ~4 = 2A -E- P -~-- P’ .

^{As in Remark}

2,

the

correspondence

such that ⁼ P’ is a linear

involution,

and let 3Q be the ruled

surface whose

generic ruling

^is

[P, S~(P)].

^{Of course it is ~ =}

.F3, and,

^{as in Remark}

2, 2)

on can see

that,

^if

Q(Pi)

⁼

Pi,

^{then the}

tangent

^to

W4

^at

Pi belongs

^{to 3t’ =}

!F3. Now,

^{as it is well}

known, through A

there passes ^a trisecant / of

W4 (intersecting

^{it at three}

non

necessarily

^distinct

points)

^which

presently

^{must be different}

from d,

^{so that ~‘ is} transversal to

W4

^at

.A,

and must then meet

W4

at two

(distinct

^or

coincident) points

^{different from .A..t then}

belongs

to F ⁼

3;~.

^{This means that}

through

A there passes ^a

ruling

^of

!F3

different from d.

On the other

hand, by

^Lemma

11,

^we have that

through

any

point B, ^{B ~} A,

^{of .~} there passes ^a

ruling

^of

9i’3

^{different from d.}

This is inconsistent with

:F3 being

^a cubic ruled surface of

Cayley’s type,

^{for which there exists}

(exactly)

^one

point

D of .~ such that the

only

^{s.l. of}

^9i’3 through

^{D is .~.}

PROPOSITION 4. In P3 let

~~

^{be a non} s.r.q.c., and let

:F3

^{be a cubic}

ruted

surface. :F3 of Cayley’s type containing ~4’

^and

having

^{its double}

base s.l. d which is a

simple

^chord

of ~4’ intersecting

^it

transversally just

^{at two distinct}

points

A and B. Then

’C4

^{cannot be s.c.t.i. on}

:F3.

PROOF. From Lemma 10 we know that the two s.l.

/~, IB,

^which

are

tangent

^to

W4 respectively

^{at A and}

B,

^are

coplanar. Through

^A

there passes ^a

(unique) trisecant 4A

^of

~4, meeting W4

^{either at three} distinct

points,

^{or at two distinct}

points being

^at

(only)

^{one of them}

tangent

^to

W4,

^or

finally tangent

^to

~4

^{at one of its}

(possible)

^flexes.

Of course

because,

^{on the}

contrary,

^the

plane

^{II =}

.tB]

would cut on

~4

^at least three

(distinct

^or

variously coincident) points laying

^{on and in at least two}

(coincident) points

^at

B,

^which

would

imply

^that absurd. Same

argument

for the trisecant of

~4 through

^{B. It is} also clear that 4A, and I meet trans-

versally ~4 respectively

^at .A and at B. This

implies that ’tA and t,,

both meet beside at A and at B

respectively,

^at two further

(distinct

^or

coincident) points.

Now the surface --Y introduced in Remark 2

presently

^coincides

with

our ~ 3, and ’tA and belong

^to

:F3.

^{Let us choose an affine}

~i3 in such a way that the

improper plane Ih

^is

hyperosculating ^W4

at one of its

(at

^{least two and}

distinct,

^{see Remark}

1) stationary points P 00;

moreover we can even suppose that one, let it be Jt A, of the two

trisecants rA and 4.,,

^{is such}

that )tA

ⁿ

C4

^{c A3.}

By

^this

W(4" =

=

W4

^{r1 A3 is a}

polynomial

^curve

~4’ _ (a(t), ^b(t), c(t))

^with

a(t), b(t),

c(t)

^E

k[t],

^{and of}

degree

^{4. Even}

~3’ - ^~3

^{r1 A3 can be repre-}

sented

parametrically

^{in the form}

(t,

^u

parameters) ,

^y

where

m(t), n(t)

^E y and can be

supposed coprime.

For every

given

^{t s.t.}

P(t)

⁼

(a(t), b(t), c(t)) ~ .~,

^the

equations (t) represent

^the

affine of the

unique

^of

through P (t ) .

^Let

t,

be a value of the

parameters

^{for which}

P (tl ) fC4,

^{but ~ A .}

The

equation (* ),

^{for t =}

tl, represents

^{the affine}

part

^{of a s.l.}

through P(t1 )

^{contained in since}

through P(tl )

^{there exists}

only

^one

ruling

of and

through P(ti)

^there passes ~~ ^c

~’3,

^then

~{t1) _

^4A-

Now let us suppose that there exists a surface # such that -~F- 9

-

3m~4. ~

^has

necessarily degree

^{4m. Given a}

generic point P(t)

of

~4’,

^{let be}

Q(P(t)), Q being

^{the linear involution}

introduced in Remark

2,

^and

.A(t), B(t)

^are

indipendent

^linear

poly-

nomials. Let

G(X, Y, Z)

^{= 0 the}

equation

^of

^{~a’ - ~}

^{r1 A3. For}

the

generic ruling g(t)

^of

^!Fa

^{we have}

where

ro(t)

⁼

A(t) jB(t). Introducing (*)

ⁱⁿ

G(X, Y, Z),

^we

get

ⁱⁿ

k[t, u]

^the

following

with suitable

because for u ⁼ 0 in

(*)

^we

get P(t).

On the other hand the

equation

has a

unique

^root

in u,

^because

4,(t)

intersects

~4 just

^at

P(t)

^and

This

implies

where

C(t)

^and

D(t)

^{are suitable}

coprime

elements in

k[t].

Let us prove ^that

D(ti)

^{~ 0.}

Suppose

^the

contrary. Now,

^of

course we have

and, by multiplying by D(t),

^we

get

By

^our

assumption

^{that I we have that}

ro(t1)

^{is defined at}

ti,

(that

^{is because}

Q(P(t1»)

^{is a}

^point

^which

^belongs

^{to ZA I}

and to the infinite value of the

parameter

^there

corresponds on C4

the

point ^{P 00 .} Putting

ⁱⁿ

( * * * )

^{t =}

t1,

^we

get

^then

whence

O(t1)

⁼

0, being Z(t), m(t), n(t) coprime:

^{absurd. So}

D(tl)

^{~ 0.}

We

get

^then

Let us assume that _~ 0. Now we

distinguish

^two cases, ^ac-

cording

^as

0(t,)

^{~ 0 or}

0(t,)

⁼ 0. In the first _{Case ’tA}

meets #,

^beside

at

A,

^{at the two} further distinct

points,

both different from

A, P(t1)

and

P(ro(t¡»,

^which

correspond

^{to the two values}

0, ^and O(t1)/D(t1)

of the

parameter u

^{in the}

parametrization

^of

~(t)~a’.

^From

^(***)

^we

have

But ’tA

intersects # also at

A,

^so

that 4,

^{c ~ because}

So t,

^{c ~ r1}

!F3

⁼

~4 :

^absurd.

In the second case we have

already n ~A~

^-

4m,

^and

P(t1)

^=A

A,

hence the same conclusion.

So it must be

Q’m(t1)

⁼

0,

^{and this}

implies again

^{that ~~ c}

~,

which leads once more to a contradiction with the

hypothesis

LEMMA 13. In P3 let

F3

^{be cubic ruled}

surface of Cayley’s type containing

^{a non} s.r.q.c.

~4’

^{let 9, be the}

unique (non singular) quadric containing ~4’

and let d be the double base s.l.

of !F3.

^{Let us} suppose de

9"

^{and let P be a}

point of ~4 (1 d,

^{at which}

~4

^{and d meet each other}

transversally.

^{Then the}

uniplanar

^double

point

^D

of ~3

^{on .d is}

dif- f erent f rom

^P.

PROOF. Within a linear

isomorphism

of P3 and with a suitable choice of an affine A3 c

P3,

^{we can} suppose that

and

by

consequence ^we have D ⁼

(0, 0, 0),

^_

fX

^{= Z =}

0}.

^Since

~ c

~,

^{we have}

Let us suppose that P ⁼ D ⁼ 0 ⁼

(0, 0).

From direct calculation it follows that

~4

passes

through

⁰

only

^{if d =} 0. On the other hand it must be b #

0,

^{otherwise no}

component

^of

~’3

m 9 would be a

quartic

curve, ^as it is easy ^{to see.} Then from the

parametric

repre- sentation of

C4

^{it is immediate to}

verify

^that

~4

^is

tangent

^{to z£}

at 0 : contradiction. So

D =1=

P.

14. In P3 let d be ^a trisecant

of

^{a non} s.r.q.c.

C4, meeting

it at three distinct

points Pi (i

⁼

1, 2, 3),

^{and let}

~’3

^{a cubic ruled}

surface of Cayley’s type having d

^{as its} double base s.l.. Then on at

least one

of

^{the f}

(i

⁼

1, 2, 3 ) of ~3, respectively through P, (i

⁼

1, 2, 3),

^{there is a}

point

^{R E with}

B Od.

PROOF.

By

Lemma 13 it is D =1=

Pi (i

⁼

1, 2, 3)

^so

through

^each

of the

points Pi,

there passes ^a

ruling

gi, with d

(i

⁼

1, 2, 3).

If on every oi

(i =1, 2, 3)

there did not exist a

point

^{.R E}

^~4

^f1 gi,

.~,

then the three distinct

planes

^{a~ =}

[~_, ~],

^{would not meet}

^ref,

outside d.

On the other hand

there

are

exactly

^three

planes through d behaving

^like

this,

^and

they

^{are the}

planes containing respectively

the three

tangents

^{to at}

Pi (i

⁼

1, 2, 3).

For any other

plane

^II

through

^{.~ we have}

~4

^r1

II)

^{= 1}

(i

⁼

1, 2, 3),

^so that 11 and

must meet each other at a further

point P 0

^d.

From all this it would follow that ^no

plane through d

^{can inter-}

sect ^on

F3 only

^d:

contradiction,

^because

F3

^{is of}

Cayley’s type.

LEMMA 15. Let

C4

^{be a non} s.r.q.c. ⁱⁿ

P3,

^{and let d be a}

tangent-

chord

of C4, tangent

^{to it at}

PI,

^and

intersecting

^it

transversally

^at

Pl

^Let

!F3

^{be a cubic ruled}

snr f ace of Cayley’s type containing ~4

and

having

^{.~ as its double base}

s.t. ;

^{moreover let us} suppose that the

uniplanar

^double

point

^D

of ~’3

^{on d is}

different from Pl.

^{Then on at}

least one

of

^the

rulings

/Ii’

(i

⁼

1, 2) of ~’3 respectively through P, (i = 1, 2)

^{there is a}

point

^{R E}

~4’

^{with d.}

PROOF.

Through

each of the

Pi (i

⁼

1, 2)

there passes ^a

ruling

_oi

different from .~: as for

P,,

^since

^{D 0} PI,

^{and as for}

Pa, by

^{Lemma 13.}

After that the

argument

^{is an} easy

adaptation

^{of that used in}

proving

Lemma 14.

PROPOSITION 5. In P3 let

~4

^{be a non} 8.r.q.c., and let

~’~

^{be a cubic}

ruled

sur f ace of Cayley’s type containing

and s2cch that its double base is:

1)

^{either a trisecant}

of rc4,

^{at three distinct}

points Pi (i

⁼

1, 2, 3);

2)

^{or a}

tangent-chord of ~4’ tangent

^to

t:t’4,

^at

P,,

^and

intersecting

it

transversally

^at

3)

or,

if ref. actually

^{has a}

flex,

^a

tangent

^{to at one}

of

^its

flexes Po .

Then,

in all these cases,

rc4,

^{cannot be s.t.c.i. on}

PROOF. Case

1 ). By

^Lemma

14, through

^{at least one of the}

points P, (i

⁼

1, 2, 3)

there passes ^a

ruling g

^of

^!F3

^{different from}

^d,

having

^{on it}

(at least)

^{two distinct}

points

^of

~4.

^{We can then} argue